1
APSC 132 PROBLEM SET 2 SOLUTIONS
Problem 1
The expansion is
adiabatic
and
irreversible
.
P
gas
V
gas
V
1
=
V
2
=
n = 1.00 mol
n = 1.00 mol
P
1
= 5050 kPa
P
2
= 101 kPa
T
1
= 773 K
T
2
=
C
v
= C
p
– R = 21.0 J/mol.K
V
1
= nRT
1
/P
1
= (1.00)(8.31)(773)/5050 L = 1.27 L
Because
process
is
irreversible
,
we
cannot
use
P
1
V
1
g
=P
2
V
2
g
.
However, we still require T
2
.
Because process adiabatic,
D
E = w (because q = 0).
Therefore, nC
v
(T
2
T
1
) = P
ext
(V
2
V
1
)
= P
ext
(nRT
2
/P
2
– nRT
1
/P
1
)
=nRP
ext
(T
2
/P
2
– T
1
/P
1
)
Unless stated otherwise, you can assume that P
ext
= P
2
.
We know
all variable except T
2
.
Solve for T
2
.
(1.00)(21.0)(T
2
– 773) = (1.00)(8.31)(101)(T
2
/101 – 773/5050)
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2
T
2
= 558 K,
D
T = 558 K – 773 K = 215 K
w =
D
E = nC
v
D
T = (1.00)(21.0)(215) J = 4515 J
D
H = nC
p
D
T = (1.00)(29.3)(215) J = 6300 J
q = 0 J
Problem 2
a)
2
isothermal expansion
P
gas
adiabatic compression
isobaric compression
1
3
V
gas
b) What do we know?
P
1
= 101 kPa
P
2
=
P
3
= 101 kPa
V
1
=
V2 =
V
3
=
T
1
= 300 K
T
2
= 400 K
T
3
= 400 K
m
1
= 2.00 kg
m
2
= 2.00 kg
m
3
= 2.00 kg
Note
: because “C” and “R” values have mass units in the
denominator, we will use masses instead of moles in the equations.
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 Fall '09
 ProfessorNewstead
 Adiabatic process, Isothermal process, Polytropic process, Isobaric process

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